Question: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $z \neq 0$. $x = \dfrac{z + 9}{-2z^3 - 28z^2 - 80z} \times \dfrac{z^3 + 18z^2 + 80z}{5z + 40} $
Explanation: First factor out any common factors. $x = \dfrac{z + 9}{-2z(z^2 + 14z + 40)} \times \dfrac{z(z^2 + 18z + 80)}{5(z + 8)} $ Then factor the quadratic expressions. $x = \dfrac {z + 9} {-2z(z + 10)(z + 4)} \times \dfrac {z(z + 10)(z + 8)} {5(z + 8)} $ Then multiply the two numerators and multiply the two denominators. $x = \dfrac {(z + 9) \times z(z + 10)(z + 8) } { -2z(z + 10)(z + 4) \times 5(z + 8)} $ $x = \dfrac {z(z + 10)(z + 8)(z + 9)} {-10z(z + 10)(z + 4)(z + 8)} $ Notice that $(z + 10)$ and $(z + 8)$ appear in both the numerator and denominator so we can cancel them. $x = \dfrac {z\cancel{(z + 10)}(z + 8)(z + 9)} {-10z\cancel{(z + 10)}(z + 4)(z + 8)} $ We are dividing by $z + 10$ , so $z + 10 \neq 0$ Therefore, $z \neq -10$ $x = \dfrac {z\cancel{(z + 10)}\cancel{(z + 8)}(z + 9)} {-10z\cancel{(z + 10)}(z + 4)\cancel{(z + 8)}} $ We are dividing by $z + 8$ , so $z + 8 \neq 0$ Therefore, $z \neq -8$ $x = \dfrac {z(z + 9)} {-10z(z + 4)} $ $ x = \dfrac{-(z + 9)}{10(z + 4)}; z \neq -10; z \neq -8 $